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\title{Chapter 05: Modules Over the Weyl Algebra}
\author{SCC ET AL}
%\institute[XX大学]{XX大学\quad 数学与统计学院\quad 数学与应用数学专业}
%\date{2025年10月}

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% 封面页
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% 目录页
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% Section 0
%\section{INTRO.}
\begin{frame}{intro. }
    
This chapter collects a number of important examples of modules over the Weyl algebra. 

The prototype of all the examples we discuss here is the polynomial ring in $n$ variables; and with it we shall begin. 

The reader is expected to be familiar with the basic notions of module theory, as explained in [Cohn 84, Ch.10].

\end{frame}

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% Section 1
\section{The Polynomial Ring} %%1
\begin{frame}[allowframebreaks]{A. }

\vspace{-0.4cm}

In Ch. 1, the Weyl algebra was constructed as a subring of an endomorphism ring. 

Writing $K[X]$ for the polynomial ring $K[x_1,\ldots,x_n]$ we have that $A_n(K)$ is a subring of $\text{End}_K K[X]$. 

One deduces from this that the polynomial ring is a left $A_n$-module. 

Thus the action of $x_i$ on $K[X]$ is by straightforward multiplication; whilst $\partial_i$ acts by differentiation with respect to $x_i$. 

This is a very important example, and we shall study it in some detail.

%---------------------------------------------------

Let us first recall some basic definitions. 

Let $R$ be a ring. 

An $R$-module is {\color{red}irreducible}, or {\color{red}simple}, if it has no proper submodules. 

Let $M$ be a left $R$-module. 

An element $u \in M$ is a {\color{red}torsion element} if $\text{ann}_R(u)$ is a non-zero left ideal. 

If every element of $M$ is torsion, then $M$ is called a {\color{red}torsion module}.

%---------------------------------------------------

\textbf{1.1. Lemma.}
Let $R$ be a ring and $M$ an irreducible left $R$-module.
\begin{enumerate}
    \item If $0 \neq u \in M$, then $M \cong R/\text{ann}_R(u)$.
    \item If $R$ is not a division ring, then $M$ is a torsion module.
\end{enumerate}

\textbf{Proof:} Consider the map $\phi : R \longrightarrow M$ defined by $\phi(1) = u$. 

It is a homomorphism of $R$-modules. 

Since $u \neq 0$ and $M$ is irreducible, $\phi$ is surjective. 

Thus (1) follows from the fact that $\ker \phi = \text{ann}_R(u)$.

Now suppose that $\text{ann}_R(u) = 0$ for some $0 \neq u \in M$. 

It follows from (1) that $M \cong R$. Since $M$ is irreducible, this can happen only if the left ideals of $R$ are trivial. 

But in this case $R$ is a division ring, contradicting the hypothesis. 

Thus $\text{ann}_R u \neq 0$, which proves (2).

Let us apply these results to the $A_n$-module $K[X]$.

%---------------------------------------------------

\textbf{1.2. Proposition.}
$K[X]$ is an irreducible, torsion $A_n$-module. 

Besides this,
$$
 K[X] \cong A_n / \sum_{1}^{n} A_n \partial_i. 
$$

\textbf{Proof:} First of all $1$ is clearly a generator of $K[X]$. 

Now suppose that $f \neq 0$ is a polynomial and consider the submodule $A_n \cdot f$. 

Let $x_1^{i_1} \ldots x_n^{i_n}$ be a monomial of maximal possible degree among the monomials that appear in $f$ with non-zero coefficients. 

Let $a$ be its coefficient. 

Thus $\partial_1^{i_1} \ldots \partial_n^{i_n} \cdot f = i_1! \ldots i_n! \cdot a$ is a non-zero constant in the submodule generated by $f$. 

Hence $A_n \cdot f = K[X]$. 

Thus $K[X]$ is irreducible. 

Since $A_n$ is not a division ring, it follows from Lemma 1.1(1) that $K[X]$ is a torsion module.

Now $1$ is a non-zero element of $K[X]$ which is annihilated by $\partial_1, \ldots, \partial_n$. 

Hence the left ideal $J$ generated by $\partial_1, \ldots, \partial_n$ is contained in $\text{ann}_{A_n}(1)$. 

Conversely, let $P \in \text{ann}_{A_n}(1)$. 

Then $P$ may be written in the form $f + Q$, where $Q \in J$ and $f \in K[X]$. 

Thus $0 = P \cdot 1 = f \cdot 1$, which implies that $f = 0$. 

Therefore $P = Q \in J$. 

We conclude that $J = \text{ann}_{A_n}(1)$. 

The isomorphism now follows from Lemma 1.1(1).

We may generalize this example as follows. 

Choose $g_1, \ldots, g_n \in K[X]$ and consider the left ideal $J$ of $A_n$ generated by $\partial_1 - g_1, \ldots, \partial_n - g_n$. 

Every element of $A_n$ is of the form $f + P$, for $f \in K[X]$ and $P \in J$; see Exercise 4.1. 

Hence the map $\psi : A_n / J \longrightarrow K[X]$ defined by $\psi(f + J) = f$ is an isomorphism of $K$-vector spaces. 

Although the action of the $x$'s is preserved under this isomorphism, it is not an isomorphism of $A_n$-modules. 

Indeed, if $f$ is a polynomial, then
$$
 \partial_i (f + J) = \frac{\partial f}{\partial x_i} + f \cdot \partial_i + J = \frac{\partial f}{\partial x_i} + f \cdot g_i + J. 
$$

Thus, $\psi(\partial_i \cdot (f + J)) = (\partial_i + g_i) \cdot f$, where the right hand side is to be calculated in $K[X]$ with its natural action. 

The module $A_n / J$ is irreducible; the proof is similar to that of Proposition 1.2.


%---------------------------------------------------

Another module that is closely related to $K[X]$ is $A_n / \sum_{1}^{n} A_n \cdot x_i$. 

As a $K$-vector space it is isomorphic to $K[\partial] = K[\partial_1, \ldots, \partial_n]$, the set of polynomials in $\partial_1, \ldots, \partial_n$. 

Using this isomorphism, we may identify the action of $A_n$ directly on $K[\partial]$: the $\partial$'s act by multiplication, whilst $x_i$ acting on $\partial_j$ gives $-\delta_{ij} \cdot 1$. 

Apart from the obvious similarities, the modules $K[\partial]$ and $K[X]$ are related in a deeper way that will be explained in the next section.

The emphasis on irreducible modules in this chapter is justified by the fact that they play the rôle of building blocks in module theory. 

However, it is not true that every $A_n$-module is a direct sum of irreducible modules. 

The sense in which we say that (interesting) $A_n$-modules are built up from irreducible modules is explained in Ch.10.

\end{frame}

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% Section 2
\section{Twisting} %%2
\begin{frame}[allowframebreaks]{B. }

\vspace{-0.4cm}

We begin with a general construction. 

Let $R$ be a ring and $M$ a left $R$-module. 

Suppose that $\sigma$ is an automorphism of $R$. 

We shall define a new left module $M_\sigma$, as follows. 

As an abelian group, $M_\sigma = M$. 

The difference lies in the action of $R$ on $M_\sigma$. 

Let $a \in R$ and $u \in M$, define $a \bullet u = \sigma(a)u$. 

A routine calculation shows that $M_\sigma$ is a left $R$-module. 

It is called the twisted module of $M$ by $\sigma$. 

Not surprisingly, $M_\sigma$ inherits many of the properties of $M$.

%---------------------------------------------------
\newpage 

\textbf{2.1. Proposition.}
Let $R$ be a ring, $M$ a left $R$-module and $\sigma$ an automorphism of $R$. 
Then:
\begin{enumerate}
    \item $M_\sigma$ is irreducible if and only if $M$ is irreducible.
    \item $M_\sigma$ is a torsion module if and only if $M$ is a torsion module.
    \item If $N$ is a submodule of $M$ then $(M/N)_\sigma \cong M_\sigma / N_\sigma$.
    \item Let $J$ be a left ideal of $R$. 
      Set $\sigma(J) = \{\sigma(r) : r \in J\}$. 
      Then $\sigma(J)$ is a left ideal of $A_n$ and $(R/J)_\sigma \cong R/\sigma^{-1}(J)$.
\end{enumerate}

\textbf{Proof:} An $R$-module $M$ is irreducible if and only if given any non-zero $u,v \in M$ there exists $a \in R$ such that $au = v$. 

This equation translates as $\sigma^{-1}(a) \bullet u = v$ in $M_\sigma$, which proves (1). 

Similarly, the equation $au = 0$ becomes $\sigma^{-1}(a) \bullet u = 0$; which proves (2). 

Now (3) is an immediate application of the first homomorphism theorem.

To prove (4), note that $\sigma(J)$ is a left ideal, since $\sigma$ is an automorphism of $A_n$. 

Let $\phi : R \longrightarrow (R/J)_\sigma$ be the homomorphism of $R$-modules defined by $\phi(1) = 1 + J$. 

If $b \in R$, then
$$
 \phi(b) = b \bullet \phi(1) = \sigma(b) + J. 
$$

Hence $\phi$ is surjective, and its kernel is $\sigma^{-1}(J)$. 

Thus (4) also follows from the first homomorphism theorem.


%---------------------------------------------------

Let us apply this construction to $A_n$. 

An important example is the Fourier transform. 

Let $\mathcal{F}$ be the automorphism of $A_n$ defined by $\mathcal{F}(x_i) = \partial_i$ and $\mathcal{F}(\partial_i) = -x_i$; see Exercise 1.4.8. 

Let $M$ be a left $A_n$-module. 

The twisted module $M_{\mathcal{F}}$ is called the Fourier transform of $M$. 

The reason for the name is clear, $\mathcal{F}$ transforms a differential operator with constant coefficients into a polynomial.

\textbf{2.2. Proposition. }
The Fourier transform of $K[X]$ is $K[\partial]$.

\textbf{Proof:} It follows from Proposition 1.2 that $K[X] \cong A_n / J$, where $J = \sum_{i=1}^{n} A_n \cdot \partial_i$. 

Since $\mathcal{F}^{-1}(J) = \sum_{i=1}^{n} A_n \cdot x_i$ we may apply Proposition 2.1(4) to get the desired result.

It follows from Propositions 2.1(1) and 2.2 that $K[\partial]$ is irreducible. 

The Fourier transform will reappear later on in this book. 

Let us consider other examples of twisting. 

For $i = 1, \ldots, n$, let $g_i \in K[x_i]$. 

Note that $[\partial_i + g_i, \partial_j + g_j] = 0$, for $1 \leq i,j \leq n$. 

Hence there exists an automorphism $\sigma$ of $A_n$ which maps $x_i$ to itself and $\partial_i$ to $\partial_i + g_i$. 

A straightforward calculation shows that
$$
 K[X]_\sigma \cong A_n / \sum_{i=1}^{n} (A_n (\partial_i - g_i)), 
$$

which is a particular case of the example considered in §1. 

Note that $K[X]_\sigma$ is irreducible, by Proposition 1.2 and Proposition 2.1(1). 

In fact $K[X]_\sigma$ is irreducible for every automorphism $\sigma$ of $A_n$.


%---------------------------------------------------

We may use the above construction to produce an infinite family of non-isomorphic irreducible left $A_n$-modules. 

For every positive integer $r$ let $\sigma_r$ be the automorphism of $A_n$ which satisfies $\sigma_r(x_i) = x_i$ and $\sigma_r(\partial_i) = \partial_i - x_i^r$.

\textbf{2.3. Theorem.}
The modules $K[X]_{\sigma_r}$ form an infinite family of pairwise non-isomorphic irreducible modules over $A_n$.

\textbf{Proof:} Let $r < t$, and suppose that there exists an isomorphism, $\phi : K[X]_{\sigma_r} \longrightarrow K[X]_{\sigma_t}$. 

Since $K[X]_{\sigma_r}$ is irreducible, it is generated by $1$. Thus $\phi$ is completely determined by the image of $1$; say $\phi(1) = f \neq 0$. 

Now the the equation $\phi(\partial_i \bullet 1) = \partial_i \bullet \phi(1)$ translates as the differential equation
$$
 \frac{\partial f}{\partial x_i} = (x_i^t - x_i^r)f. 
$$


The left hand side of the equation has degree $\leq \deg f - 1$. 

Since $f \neq 0$ and $r < t$, the right hand side has degree $\deg f + t$. 

This is a contradiction, so the theorem is proved.

The construction of irreducible $A_n$-modules leads to many important questions and we shall return to it several times.


\end{frame}

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% Section 3
\section{Holomorphic Functions} %%3
\begin{frame}[allowframebreaks]{C. }

\vspace{-0.4cm}

Let $U$ be an open subset of $\mathbb{C}$. 

The set $\mathcal{H}(U)$ of holomorphic functions defined on $U$ contains the polynomial ring $\mathbb{C}[z]$. 

We will make it into a left $A_1(\mathbb{C})$-module. 

Denote the generators of $A_1(\mathbb{C})$ by $z$ and $\partial = d/dz$. 

Then $z$ acts by multiplication, whilst $\partial$ acts by differentiation on $\mathcal{H}(U)$. 

It is routine to check that the actions are well-defined and satisfy the required properties; see Appendix 1. 

As an $A_1(\mathbb{C})$-module, $\mathcal{H}(U)$ is not irreducible. 

This is the same as saying that there are holomorphic functions which are not polynomials. 

On the other hand, a torsion element of $\mathcal{H}(U)$ is a holomorphic function which satisfies an ordinary differential equation with polynomial coefficients. 

These elements exist. 

For example, $\exp(z)$ is a solution of $d/dz - 1$. 

However, $\mathcal{H}(U)$ is not a torsion module. 

This is a more interesting result that we now prove. 

In fact we show that the function $\exp(\exp(z))$ does not satisfy a polynomial differential equation. 

First a technical lemma.

\textbf{3.1. Lemma.}
Let $h(z)$ be the holomorphic function $\exp(\exp(z))$. 

For every positive integer $m$ there exists a polynomial $F_m(x) \in \mathbb{C}[x]$ of degree $m$ such that
$$
 d^m h / dz^m = F_m(e^z) h(z). 
$$


\textbf{Proof:} Since $dh/dz = e^z \cdot h(z)$, the result holds for $m=1$ with $F_1(z) = z$. 

Suppose, by induction, that the formula holds for $m$. 

Using the formula $d^{m+1} h / dz^{m+1} = d(d^m h / dz^m) / dz$ and the induction hypothesis, we have that
$$
 d^{m+1} h / dz^{m+1} = F'_m(e^z) h(z) e^z + F_m(e^z) F_1(e^z) h(z), 
$$



where $F'_m$ is the derivative of the polynomial $F_m$. 

Putting
$$
 F_{m+1}(z) = F'_m(z) z + F_m(z) F_1(z) 
$$

we have the desired formula. 

Besides this, the degree of $F_{m+1}$ is $\deg F_m + 1 = m + 1$, by the induction hypothesis.


%---------------------------------------------------

In the proof of the next proposition we use the fact that the exponential function is not algebraic. 

A complex function $f \in \mathcal{H}(U)$ is algebraic if there exists a non-zero polynomial $G(x,y) \in \mathbb{C}[x,y]$ such that $G(z,f(z)) = 0$, for all $z \in U$. 

For a proof that $e^z$ is not algebraic, see Hardy's delightful book {\color{red}The integration of functions of a single variable}, [Hardy 28, Ch.V, §16].

\textbf{3.2. Proposition.}
The function $h(z) = \exp(\exp(z))$ is not a torsion element of the $A_1(\mathbb{C})$-module $\mathcal{H}(U)$.

\textbf{Proof:} Let $P = \sum_{i=0}^{r} f_i(z) \partial^i$ be an element of $A_1(\mathbb{C})$. 

We assume that $f_r \neq 0$. 

Then $P \cdot h = 0$ in $\mathcal{H}(U)$ is equivalent, by definition, to the differential equation
$$
 \sum_{i=0}^{r} f_i(z) \frac{d^i h}{dz^i} = 0. 
$$


Using Lemma 3.1 and the fact that $h(z) = \exp(\exp(z)) \neq 0$ for every $z \in \mathbb{C}$, we obtain an equation of the form
$$
 \sum_{i=0}^{r} f_i(z) F_i(e^z) = 0. 
$$

This is equivalent to $G(z,e^z) = 0$, where $G(x,y) = \sum_{i=0}^{r} f_i(x) F_i(y)$ is a polynomial in two variables. 

Since $F_i(y)$ is a polynomial in $y$ of degree $i$, the polynomial $G(x,y)$ must be non-zero. 

This implies that $e^z$ is an algebraic function; a contradiction.

In the next chapter, $\mathcal{H}(U)$ will play the rôle of module of solutions for differential equations represented in $\mathcal{D}$-module language.

%---------------------------------------------------

\end{frame}

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% Section 4
\section{Exercises}
\begin{frame}[allowframebreaks]{D. Exercise. }

\vspace{-0.4cm}

\textbf{Exercise 4.1.} Let $g_1, \ldots, g_n$ be polynomials in $K[X]$.
\begin{enumerate}
\item Show, by induction on $m$, that $\partial_i^m$ may be written in the form $D(\partial_i - g_i) + f$, where $D \in A_n$ and $f \in K[X]$.
\item Conclude that every element of $A_n$ can be put in the form $Q + f$, where $Q \in \sum_{1}^{n} A_n (\partial_i - g_i)$ and $f \in K[X]$.
\end{enumerate}

\newpage 

\textbf{Exercise 4.2.} Show that if $g_i \in K[x_i]$ then the module $A_n / \sum_{i=1}^{n} (A_n (\partial_i - g_i))$ is irreducible.

\newpage 

\textbf{Exercise 4.3.} A left $R$-module $M$ is cyclic if $M = R \cdot u$, for some $u \in M$. 

Show that an irreducible module is always cyclic.

\newpage 

\textbf{Exercise 4.4.} Let $R$ be a simple ring that is not a division ring. 

Suppose that $M, M'$ are non-zero left torsion $R$-modules. 

Show that if $M$ is cyclic and $M'$ is irreducible, then $M \oplus M'$ is cyclic.

Hint: Let $u$ be a generator of $M$. 
Choose $0 \neq a \in R$ such that $au = 0$. 
Since $R$ is simple $aM' \neq 0$. 
Let $v \in M'$ with $av \neq 0$. 
Then $u + v$ generates $M \oplus M'$.

\newpage 

\textbf{Exercise 4.5.} Using Exercise  4.4, show that the direct sum of any finite number of irreducible $A_n$-modules is cyclic.

\newpage 

\textbf{Exercise 4.6.} Let $M, M'$ be left $R$-modules and let $\sigma$ be an automorphism of $R$. 

Show that $(M \oplus M')_\sigma \cong M_\sigma \oplus M'_\sigma$.

\newpage 

\textbf{Exercise 4.7.} Show that $\mathcal{H}(U)$ is not a cyclic $A_1(\mathbb{C})$-module.

Hint: Let $h \in \mathcal{H}(U)$ be a generator. 

Show that $h$ cannot be constant and that $\exp(h) \notin A_1(\mathbb{C}) \cdot h$.

\newpage 

\textbf{Exercise 4.8.} Let $U$ be an open set of $\mathbb{R}^n$ and $C^\infty(U)$ be the real vector space of all functions of class $C^\infty$ defined on $U$. 

Let $x_i$ act by multiplication and $\partial_i$ by differentiation on $C^\infty(U)$. 

Show that this makes $C^\infty(U)$ into a left $A_n(\mathbb{R})$-module. 

Is $C^\infty(U)$ an irreducible $A_n(\mathbb{R})$-module? Is it a torsion module? Is it cyclic?

\newpage 

\textbf{Exercise 4.9.} Are $\sin(e^z)$ and $\cos(e^z)$ torsion elements of the $A_1(\mathbb{C})$-module $\mathcal{H}(\mathbb{C})$?

\newpage 

\textbf{Exercise 4.10.} Are $\exp(\sin z)$ and $\exp(\cos z)$ torsion elements of the $A_1(\mathbb{C})$-module $\mathcal{H}(\mathbb{C})$?

\newpage 

\textbf{Exercise 4.11.} Show that $\cos z$ and $\sin z$ are torsion elements of $\mathcal{H}(\mathbb{C})$.

\newpage 

\textbf{Exercise 4.12.} Let $U$ be the open set $\mathbb{C} - (-\infty, 0]$. 

If $\alpha \in \mathbb{R}$, show that $z^\alpha$ is a torsion element of the $A_1(\mathbb{C})$-module $\mathcal{H}(U)$.

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